1
$\begingroup$

How to find the limit of:

$\lim_{n\to\infty} \sqrt[n]{n+\sin^2n}$

using squeeze theorem?
Because $0\le \sin^2n \le 1 $, I find $a_n=\sqrt[n]{n}$ (which is equal to 1) and $c_n=\sqrt[n]{n+1}$, but I don't know how to prove that second formula is also a 1. Could someone help me solve it please? Thank you!

$\endgroup$
  • $\begingroup$ How do you know $\lim \sqrt[n]{n}=n$? You can't just assume. The proof of $\lim \sqrt[n]{n+1}$ isn't much different. $\endgroup$ – fleablood Feb 6 at 19:32
  • $\begingroup$ If you know that $2^{1/n} \rightarrow 1$ and $n^{1/n} \rightarrow 1,$ then you can use the fact that $\sqrt[n]{n} \leq \sqrt[n]{n + {\sin}^2 n} \leq \sqrt[n]{n+n} = 2^{1/n}n^{1/n},$ and note that the left side approaches $1$ and the right side approaches $1 \cdot 1 = 1.$ $\endgroup$ – Dave L. Renfro Feb 6 at 19:34
  • $\begingroup$ $n^{1/n} \lt (n+1)^{1/n} \lt (2n)^{1/n}=2^{1/n}n^{1/n}.$ $\endgroup$ – Peter Szilas Feb 6 at 19:34
  • $\begingroup$ Thank you both! Easier than I thought $\endgroup$ – L.krez Feb 6 at 19:42
1
$\begingroup$

Here's a hint:

$$n<n+1\leq 2n\implies \sqrt[n]{n}<\sqrt[n]{n+1}\leq \sqrt[n]{2n}=\sqrt[n]{2}\cdot\sqrt[n]{n}.$$

$\endgroup$
  • $\begingroup$ (+1) Exactly what I gave in a comment, but your answer came first! $\endgroup$ – Dave L. Renfro Feb 6 at 19:35
  • $\begingroup$ Thank you, now it seems obvious! $\endgroup$ – L.krez Feb 6 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.